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~[4 pointsUse the integral test to determine whether the seriesn(In(n))2 converges or diverges_lim b-0dx x(In(x))2lim b -By the Integral Test, the infinite series c...

Question

~[4 pointsUse the integral test to determine whether the seriesn(In(n))2 converges or diverges_lim b-0dx x(In(x))2lim b -By the Integral Test, the infinite series converges because the improper integral converges_ By the Integral Test, the infinite series diverges because the improper integral diverges _-/4 pointsUse the integral test to determine whether the series converges or diverges n2 _ 9 n = 4lim b-0ax x2 _ 9lim b-0By the Integral Test, the infinite series converges because the improper i

~[4 points Use the integral test to determine whether the series n(In(n))2 converges or diverges_ lim b-0 dx x(In(x))2 lim b - By the Integral Test, the infinite series converges because the improper integral converges_ By the Integral Test, the infinite series diverges because the improper integral diverges _ -/4 points Use the integral test to determine whether the series converges or diverges n2 _ 9 n = 4 lim b-0 ax x2 _ 9 lim b-0 By the Integral Test, the infinite series converges because the improper integral converges: By the Integral Test, the infinite series diverges because the improper integral diverges_



Answers

Determine whether each improper integral converges or diverges. If it converges, find its value. $$ \int_{-\infty}^{\infty} \frac{d x}{x^{2}+4} $$

And they were given stories on the farm One of a square rhythm and rest far goes from one infinity to use the Interco test and we consider the integral on the form in the government wants infinity one divided by the square root on the express for the X doesn't weaken Written you can read that as X plus for off a minus one out of two from one to infinity The X there found the entire the reality of this one week O Judah blast Ah, about off 1/2 and dividing by 1/2 hand will be given Everyone comes to And we were very entirely riveting from one to infinity So it we would infinity inside Uganda infinity and then minus ah far to times square root I'm the five divided by one So stagecoach infinity there far by the Interco test here we can conclude that miseries here will be diversion to

Can discussion a series is given that is submission and it calls to 12 in finite and divided by any square plus one. Okay. And now we have to decide whether the series converges or diverges using that integral test. Okay? So first of all I am in the cities is and divided by any square plus one. Okay. And two integral test. First of all, we will take the function F X equals two. If X equals two X divided by X square plus one. Okay? But before applying the integral test we have to check whether the series is decreasing and positive or not. Okay? So for a series decreasing the submission equals and equals to one point finite. Am should be equals to zero. Okay? And the series is positive we can say and 4°.. We will test this. So it will be integration and equals to 12 in finite and A N. S. And divided by any square plus one. Okay. And when we saw it will be integration and equals to one in finite. And we will take and common factor so it will be one divided by and bless one by n. Okay. And when we apply the limit here this will be zero and it will be in finite and one by in finite that will be zero. Okay. So we can say our series is decreasing and it is positive. So it is safe to apply the integral test here. Okay. And now our function is X divided by X squared plus one. So for integral test integration of F X D X. That is X divided by X squared plus one dx from one to in finite. Okay. We have to check this F And now in this integral test three and 9.3. Okay. If this integral converges so our series also converges and if this integral diverges over series also diverges. Okay, so we will solve this integration So it will be okay integration of here 112 in finite. So the integration of X divided by X square plus one. It will be Ln one by two. L n x square plus one. Okay. And the limit is here +12 in finite. Okay. And now when we apply this it will be mm one by two. Okay. First a parliament that will be in finite and minus lower element. It will be won by two. Ln two. Okay. Or we can say this will be infinite minus some value. That will be in my night. So here this integral diverges. Okay, so we can say the integral diverges over series will also diverges. Okay, so I'm writing down the final answer here diverges and this will be the final answer of discussion. Thank you.

We want to use the interval test to determine whether this series converges or diverges. They tell us to check the conditions to satisfy integral test. So let's go ahead and check the conditions first. So if we let f be eaten too So one thing that we can actually do is just go ahead and factor out this seven one of the properties of the Siri's. And we can just look at one over the spirit of end plus four because if this convergence that multiplying it by seven also converges so no issues with that and I'm just gonna do this so we don't have to write out as much. So we have one over and plus four square root. Now, this is always going to be greater than zero. Because if we add so we're starting from and is he gonna one? And they were going to infinity. So we add some positive. Number 24 Still positive square root Positive number, positive ID by positive number Positive. So this is always going to be good. Next, we need to check to see if this is continuous and well, this is going to be continuous as long as our denominator is not equal to zero as well as we're not plugging in any negative numbers. So we could just do and plus four greater than zero subtracted over That says we need to ensure that in is strictly greater than negative foreign. Well, we're starting from and is a good one. So that checks out. And we also need to see if it is decreasing so we can check our derivative. So remember, this is really going to be in to the negative 1/2 power. So using power rule that says we get negative 1/2 times and plus four to the negative 3/2. So then we'd use chain rule, take drift on the inside, so we just get one. So rewriting this, we would have negative one over to times in plus four to the three halfs power. And again, if we're plugging in values larger than one that are denominators going to be positive. But then we're dividing into a negative number. So this is going to be less than zero for all values that this is actually defined for. So what looks like this checks out so we could go ahead and apply the integral test to this here. All right. And so I'm going to go ahead and rewrite this inter, although so will be one to infinity. Oh, in plus four raised to the negative 1/2 power. And I want to do this because then it will be a little bit easier to see that we could use a use substitution to integrate this. So I'm going to let you be in plus four. And so then that's going to give us that d U is equal to D in. Let's go ahead and plug all of that in and our new bounds. Let's see what those are. Well, we plug one in for end, I guess us five plugging infinity or we take the limit as in the purchase Infinity. So you is still going to go to infinity and over here we're gonna have you to the negative 1/2 dia burnt. Now we could go ahead, imply power works. Interview this. So this should be to you to the 1/2 which something is the square root of you and we evaluate from five to infinity. So now if we take the limit as u approaches infinity of the square root of you. Well, that's going to diverge and then plug it in five just gives us to route by. So this integral here diverges. So that implies that are Siri's of seven over the square root of in squared plus four. And is he goto one to infinity diverges as well.

Using the integral test of the Siri's. We'll take the limit a Z. He goes to infinity of the integral off one from one to be. And then we're going to use the power rule on the outside here. So this would give us a new power of N plus four to the one half and then divided by one half of multiplied by two. Then we don't need to worry about you substitution, since the derivative of N Plus four is just one. So I would just divide that whole expression by one. And then we're going to evaluate this from one to be. And of course, we're still letting the approach infinity. And so therefore, this would leave us with infinity in the mix here. And of course, that that's going to diverge off to infinity. And it doesn't matter what the rest of it is. So therefore, the original integral or excuse me. The original series is divergent by the integral tests since the integral diverged off to infinity


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